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Thread: New Sniper install. Fuel issue, crank issue & direction question.

  1. #11

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    Quote Originally Posted by SatherCS View Post
    I know the coil wire has two wires so that there 's enough voltage during cranking to fire the original ignition, that one I believe he was using for the MSD box keyed 12V wire. If you look at wire #904, that is the wire that tells the voltage regulator to turn on and has full 12V in run & crank, because it is not running through the resistor wire like the ignition is. This is the wire I use, and is the wire I think bannind was using before he made the relay system.

    Also bannind, I thought the Sniper came with a coil driver? You can use that if you don't want to use the MSD with timing control, even if you don't go with the Dual Sync distributor. I will say one thing about the MSD or any CDI style box, it is definitely true that they will fire fouled plugs more easily than an inductive style single coil ignition. I ran into that problem a couple of weeks ago in my '91 TBI 350. I had removed my MSD to try and figure out some fuel consumption problems, and was running on the factory setup. After a couple days of just starting & moving it a couple feet, the factory ignition couldn't fire the plugs. I still had the MSD in the engine bay, so I put it back on and it fired right up (even with me flooding the engine).
    #16 & #904 are the same wire connection on the ignition switch: www.bronco.com/graphics/diagrams/68-71ignition-color.gif
    #904 "seems" like it would be a true ignition wire "only" when #262 IS connected to #16.
    Or in other words, #262 has to be connected to #904 to work right. Or #262 has to be connected to #16 to work right. #16 & #904 are connected together at the ignition switch.
    Last edited by BITE_ME; 01-31-2018 at 03:27 AM.

  2. #12

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    Everything at the ignition switch has +12V because the switch gets it power from #21, which is connected to a big splice that has #37 (Alternator wire) & #37B (Ammeter wire), which eventually connects to #38 (Ammeter to Battery side of solenoid). The power to the switch does not come from the coil. If you measure #16 at the ignition switch, it will have +12V. The electricity then flows OUT of the switch and is reduced to 6V along the length of the resistor wire.
    Last edited by SatherCS; 01-31-2018 at 03:58 AM.

  3. #13

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    Your point seems to be that power can't come from #262, and in turn feed #16.
    Look at the picture "inside" the starter solenoid. A solenoid "can" send power to more then one circuit.
    Specifically. It's sending power to the starter & #262.
    #262 & #16 "are" connected "together.
    Or to put it in the most simplest terms, to a person who has never worked on a car with a "points" ignition.
    Points ignition systems "always" use two power sources.
    1. First power source is "straight" battery power. This comes from the starter solenoid (specifically old Fords). Straight battery power is used. Because the starter draw will pull down the voltage too low, to be usable for the next power source for the points.
    2. The "run" power source. This is "less" voltage then "straight" battery power. It goes through resistance wire. This "less" voltage keeps the points/coil happy, so they last longer. Ford connects these two wires "together" to make a "true ignition" wire. Because "both" of these wires are connected, when one has "power", so does the "other".

    The only other way I can explain this more simply is.
    Remove the #262 wire "off" the solenoid. So it's not connected to anything.
    When you do this #16 & #904 (Red/Green, Green/Red) will not have battery voltage when cranking the starter.

  4. #14

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    Alright, I drew a picture. The resistor wire & voltage regulator wire are in parallel. This means that both branches of the circuit have the same voltage in them. In this case, the branch with the coil has a resistor, which only lets the coil see about 6V when the solenoid is not engaged. When the ignition switch triggers the "S" terminal of the solenoid to activate it, the "I" terminal creates another connection to the coil essentially bypassing the resistor (so that the coil can fire during starting). If you had an ammeter hooked up to in-line with the resistor wire, you'd see that there would be no current flowing through the resistor when the starter is activated, because both sides of the resistor are at battery voltage (about 10V depending on the starter). The solenoid will give power to the coil during cranking, but power will not flow back through the resistor towards the switch, which also means the wire #904 is not affected by the solenoid. Now the wiring is 50 years old at this point, so there will be some extra resistance in the wires, so you may see lower voltage at the ends of the wires because of that, and also the length of the circuit.
    Attachment 2815
    EDIT: Also, sorry for going off-topic. I think the relays were a good idea (every new car has tons of them). I probably need to do something like that as well, mainly for taking as much current off of the original wiring as possible.
    Last edited by SatherCS; 01-31-2018 at 05:45 PM.

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